Need help with this problem

Bob can cut through a log in 3 minutes. How long will it take him to cut a 25foot log into 5 foot sections?

Tom [10]

15 minutes. 25/5 = 5

5×3 = 15

4 0

3 years ago

Help! I need it bad! (question below)

lara31 [8.8K]

Answer:

Problem and solution

Step-by-step explanation:

"Struggle", "threat", and "trouble" are words that describe a problem.

"Hope", "possibility", and "future" are words that describe a solution.

8 0

3 years ago

How do I write 0.5 million dollars as a decimal...for example, 1.0 million is 1,000,000. I don't get we're you put the decimal p

Lana71 [14]

Well .5 of a million dollars is 500,000. so you are right the decimal of 500,000 is .5 million.

6 0

3 years ago

Is the distance between -7 and -13 on a number line 6?

sattari [20]

Answer:

yes

Step-by-step explanation:

13-7=6 so that's the difference

4 0

3 years ago

For a test of population proportion H0: p = 0.50, the z test statistic equals 1.05. Use 3 decimal places. (a) What is the p-valu

sergiy2304 [10]

Answer:

(a) The p-value of the test statistic is 0.147.

(b) The p-value of the test statistic is 0.294.

(c) The p-value of the test statistic is 0.8531.

(d) None of the p-values give strong evidence against the null hypothesis.

Step-by-step explanation:

The p-value is well defined as the probability,[under the null hypothesis (H₀)], of attaining a result equivalent to or greater than what was the truly observed value of the test statistic.

We reject a hypothesis if the p-value of a statistic is lower than the level of significance α.

The null hypothesis for the test of population proportion is defined as:

H₀: p = 0.50

The value of z-test statistic is,

z = 1.05

(a)

The alternate hypothesis is defined as:

Hₐ: p > 0.50

Compute the p-value of the test statistic as follows:

$p-value=P(Z>1.05)\\=1-P(Z$

*Use a z-table for the probability value.

Thus, the p-value of the test statistic is 0.147.

(b)

The alternate hypothesis is defined as:

Hₐ: p ≠ 0.50

Compute the p-value of the test statistic as follows:

$p-value=2\times P(Z>1.05)\\=2\times 0.1469\\=0.2938\\\approx 0.294$

*Use a z-table for the probability value.

Thus, the p-value of the test statistic is 0.294.

(c)

The alternate hypothesis is defined as:

Hₐ: p < 0.50

Compute the p-value of the test statistic as follows:

$p-value= P(Z1.05)=1- 0.1469\\=0.8531$

*Use a z-table for the probability value.

Thus, the p-value of the test statistic is 0.8531.

(d)

The decision rule of the test is:

If the p-value of the test is less than the significance level α, then the null hypothesis is rejected at α% level of significance.

And if the p-value of the test is more than the significance level α, then the null hypothesis is failed to be rejected.

The most commonly used level of significance are:

α = 0.01, 0.05 and 0.10

The p-value for all the three alternate hypothesis are:

p-values = 0.147, 0.294 and 0.8531.

All the p-values are quite large compared to the α values.

Thus, none of the p-values give strong evidence against the null hypothesis.

The null hypothesis was failed to be rejected.

5 0

3 years ago